Prove that if f is one-to-one, then f(x) ∩ f(y ) = f(x ∩ y ) for all x, y ⊆ a. is the converse true?

See proof below, the converse is true

Step-by-step explanation:

Let f be a function whose domain is the set A, and X,Y be arbitrary subsets of A. One way to prove the equality of sets f(X)∩f(Y)=f(X∩Y) is, by definition, to prove the inclusions f(X)∩f(Y) ⊆ f(X∩Y)  and f(X)∩f(Y )⊇f(X∩Y).

First, let a∈f(X)∩f(Y). Then a∈f(X) and a∈f(Y) by definition of intersection. Thus, a=f(t) for some t∈X (because a is an element of the direct image of X) and a=f(s) for some s∈Y. Then a=f(t)=f(s). We know that f is 1-1, so the previous equality implies that t=s. Now, t∈X and t=s∈Y, thus t∈X∩Y. Therefore, a=f(t) for some t∈X∩Y, which means that a is an element of the direct image of X∩Y, that is, a∈f(X∩Y). This holds for all a, thus f(X)∩f(Y) ⊆ f(X∩Y).

For the second inclusion, let b∈f(X∩Y), then there exists some z∈X∩Y such that b=f(z). Since z∈X∩Y, we have that z∈X and z∈Y. Thus, b=f(z) for some z∈X, that is, b∈f(X). Similarly, b∈f(Y) because b=f(z) and z∈Y. Therefore b∈f(X)∩f(Y) for all  b∈f(X∩Y), then f(X)∩f(Y )⊇f(X∩Y). Note that this holds for any function f, as we did not use that f is 1-1.

To prove the converse, suppose that f(X)∩f(Y)=f(X∩Y)  for all  X, Y⊆ A. We will prove that f is 1-1, that is, for all a,b∈A if f(a)=f(b) then a=b.

Define the sets X:={a} and Y:={b}. Then f(X)={f(a)} and f(Y)={f(b)}. Assuming that f(a)=f(b), we obtain f(X)=f(Y). We know that f(X)∩f(Y)=f(X∩Y), which means {f(a)}=f(X∩Y). Then, there exists some s∈X∩Y such that f(s)=f(a), thus X∩Y is a non-empty set, which implies that a∈Y. But Y has only one element, b, therefore a=b which shows that f is 1-1.

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step-by-step explanation:

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb